# Equilateral triangles on two adjacent sides of a rectangle, and the rectangle's fourth vertex, determine another equilateral triangle

Recently I bumped into a $$7^{th}$$-grade problem. Shamefully I can't find any elementary solution for it.

The problem is as follows:

There is given a rectangle $$ABCD$$ with shorter sides $$AD=BC$$. Let $$BCE$$, $$ABF~$$ be two equilateral triangles with $$E~$$ inside $$ABF$$. We are being asked to prove that $$DEF$$ is a equilateral triangle as well.

Any clue how to defeat it $$7th$$-graders with elementary methods?

• This approach might be a little weird for a $7$th-grader, but ... $\triangle EBF\cong\triangle ECD$ (via a bit of angle chasing, and SAS), so $\overline{EF}\cong\overline{ED}$. Since a $60^\circ$ rotation about $E$ aligns $\overline{EB}$ with $\overline{EC}$ —and, thus, one triangle with the other— we conclude that $\angle DEF=60^\circ$. The result follows. $\square$
– Blue
Nov 8, 2022 at 18:13
• Also this question isn't a duplicate according to what I have searched on Approach0, which was "two equilateral triangles in rectangle". Nov 9, 2022 at 7:31
• Treating the geometry as merely a picture of the underlying algebra, we shouldn't need $AD<AB$ (the theorem should be true regardless), and it should be true even if the triangles $ABF$ and $BCE$ are outside $ABCD$. Nov 9, 2022 at 7:44
• Ah yes I just needed the fact that $DC = AB = BF$ to make sense of Blue's comment. Nov 9, 2022 at 7:52

We have $$AD = EC,$$ $$AF = CD,$$ and $$\angle DAF = 30^\circ = \angle ECD.$$ By the side-angle-side property we therefore have congruent triangles $$\triangle DAF \cong \triangle ECD.$$ Therefore $$DF = DE.$$

To finish the proof we just need to show that $$\angle EDF = 60^\circ,$$ which I suppose we can angle-chase in a few ways. One way is, since the sum of angles of a triangle is $$180^\circ,$$ we have $$\angle CDE + \angle CED = 180^\circ - \angle DCE = 180^\circ - 30^\circ = 150^\circ.$$

Then \begin{align} \angle EDF &= \angle EDC + \angle CDF \\ &= \angle EDC + (\angle ADF - \angle ADC) \\ &= (\angle EDC + \angle ADF) - 90^\circ \\ &= (\angle EDC + \angle CED ) - 90^\circ \\ &= 150^\circ - 90^\circ \\ &= 60^\circ. \\ \end{align}

• I'd say this should be the very answer the authors of the problem had have in mind when created the problem. Nevertheless If not then this one is the excellent answer no doubts! Nov 8, 2022 at 19:43
• This answer is lovely, but I think can be incrementally improved even further. If from the start you use $\triangle EBF$ and $\triangle ECD$, then showing similarity is formally the same (SAS with $EB=EC$, $BF=CD$, $\angle EBF = 30^{\circ} = \angle ECD$), but the transformation between them is easier to visualise — rotation around the shared vertex $E$ — and leads to a simpler angle-chase in the endgame: $\angle DEC = \angle FEB$ (by the similar triangles) so $\angle DEF = \angle DEC - \angle CEF = \angle FEB - \angle CEF = \angle CEB = 60^{\circ}$. Nov 9, 2022 at 8:23
• We may also show that $\angle DAF = \angle ECD = \angle EBF = 30^o$ , $DA=EC=EB$ and $AF=CD=BF$. Hence $\Delta DAF \cong \Delta ECD \cong \Delta EBF$ (SAS) and finally $DF=ED=EF$ Nov 9, 2022 at 12:03
• @PeterLeFanuLumsdaine I’m not sure when isometries are taught these days, but that argument shows the advantage of picking your triangles carefully. I like it! Nov 10, 2022 at 4:55
• @LiKwokKeung That’s a very good argument. So we actually don’t even need to chase angles; just SAS in three places. Nov 10, 2022 at 4:56

I'm not sure if this would count as 7th grader mathematics but, there is in fact a way to prove this using ideas strictly from elementary geometry as I'll explain below:

1.) Connect point $$F$$ with $$C$$ via segment $$FC$$. Since $$\angle ABF=60$$, we know that $$\angle FBC=30$$ and that implies $$\angle FBE=30$$ as well. This proves that segment $$FB$$ is the perpendicular bisector of segment $$EC$$ and the angle bisector of $$\triangle BEC$$. It follows that $$\triangle FBE$$ and $$\triangle FBC$$ are congruent, therefore $$FC=FE$$.

2.) Since we know that $$\angle FBC=30$$, we can also conclude that $$\angle FAD=30$$ as well. We can also notice that $$\triangle FBC$$ and $$\triangle FAD$$ are also congruent via the SAS property, therefore we can also infer that $$FC=FE=FD$$. This proves that point $$F$$ is the circumcenter of $$\triangle DEC$$, therefore $$\angle DFE$$ would be twice the measure of $$\angle ECD$$. We can see that $$\angle ECD=90-60=30$$, therefore $$\angle DFE=60$$ and since $$FE=FD$$, this proves that $$\triangle DFE$$ is in fact equilateral.

Another way. Let the shorter side be $$2a$$ and the other be $$2b$$. Referred to cartesian coordinates we can made $$A=(0,2a),B=(2b,2a),C=(2b,0),D=(0,0)$$ so we have because of equilateral triangles $$\triangle{BEC}$$ and $$\triangle{AFB}$$ $$E=(2b-a\sqrt3,a) \text { and } F=(b,2a-b\sqrt3)$$ With this we have immediately $$\overline{ED}=\overline{DF}=\overline{EF}=2\sqrt{a^2+b^2-ab\sqrt3}$$

• This is a good approach but inappropriate for this answer, I believe. Since it is meant for 7th graders that are only familiar with elementary geometry Nov 8, 2022 at 16:21
• I don't know what the level of the $7^{th}$ graders is and we assume that basic cartesian coordinates were known by them. But I want to believe you. Thanks. Nov 8, 2022 at 20:28

I think I've found a good 7th-grade explanation with no equations needed. It requires drawing another equilateral triangle connecting points A and D and then comparing angles.

See illustration here

We're given that triangle AFB is equilateral. Thus, angle AFB is 60 degrees.

Notice that line BF is a perpendicular bisector of line EC. I've labeled the intersection point M in the illustration.

Because of this, FC = FE = FD.

Angle DFE must therefore also be 60 degrees, because angles DFA and EFB are equal as you can see in the diagram (triangles EFM and DFN are congruent).

Since angle DFE is 60 degrees, and DF = FE, triangle DFE must be equilateral.

Here is an approach that aims to show $$DE = EF = FD$$ to show that $$\Delta DEF$$ is equilateral.

Firstly, as $$DA = CB$$, $$AF = BF$$, and $$\angle DAF = \angle CBF = 90º - 60º$$, $$\Delta DAF$$ is congruent to $$\Delta CBF$$ and so $$FD = FC$$. Now $$\angle ABE = \angle FBC = 90º - 60º$$ and thus $$\angle ABE = \angle EBF = \angle FBC = 30º$$. By $$SAS$$ congruence again, $$\Delta FEB$$ is congruent to $$\Delta FCB$$, and hence $$FE = FC$$, which yields $$FD = FE = FC$$.

Using $$SAS$$ again on $$\Delta AEB$$ and $$\Delta DEC$$ (or the fact that $$E$$ lies on the perpendicular bisector of $$BC$$), it is not too hard to see that $$AE = ED$$. However as mentioned previously, $$\angle ABE = \angle EBF$$, and once more using $$SAS$$, $$AE = EF$$.

Thus $$AE = EF = ED$$ from above, and from before, $$FE = FD$$.

Therefore $$EF = ED = FD$$, hence proved.

• One fairly unexpected thing from this proof is that $AD = EC$, which is interesting to note. Nov 9, 2022 at 7:16
• Hm?? $AD=BC$ is given, and $BC=CE$ by construction of $E$. Nov 9, 2022 at 7:50
• Right, didn't catch that ;( Nov 9, 2022 at 7:51